Islamic University of Gaza ‐Environmental Engineering Department
Water Treatment
Water Treatment
EENV 4331
Lecture 2: Aeration
Dr. Fahid Rabah
1
2.1 Purpose of Aeration in water Treatment
Aeration is used for many purposes in water Treatment , the
main uses are:
1.
Increase the dissolved oxygen concentration in water.
2.
Decrease CO2
corrosiveness.
3.
Reduce taste and odor caused by dissolved gases such as
Hydrogen sulphide (H2S) and methane (CH4) that are
removed during aeration.
4.
Oxidize iron and manganese from their soluble to
insoluble states and cause them to precipitate so that they
maybe removed by sedimentation and filtration processes.
5.
Remove certain volatile organic compounds.
concentration
thereby
reduce
its
2
2.2 Types of aerators used in water Treatment
The most common aerators used in water treatment are :
1.
Gravity Aerators: In gravity aerators, water is allowed to fall
by gravity such that a large area of water is exposed to
atmosphere, sometimes aided by turbulence. Cascade
aerators and Multi‐tray aerators are two examples of this
type.
2.
Fountain Aerators : These are also known as spray aerators
with special nozzles to produce a fine spray.
3.
Injection or Diffused Aerators : It consists of a tank with
perforated pipes, tubes or diffuser plates, fixed at the
b tt
bottom
t release
to
l
fi air
fine
i bubbles
b bbl from
f
compressor unit.
it
4.
Mechanical Aerators : Mixing paddles as in flocculation are
used.
d Paddles
P ddl may be
b either
ith submerged
b
d or att the
th surface.
f
3
Cascade Aerator
4
Cascade Aerator
5
Cascade Aerator
6
Cascade Aerator
7
Cascade Aerator
8
Cascade Aerator
9
Cascade Aerator
10
Cascade Aerator
11
Multi Tray Aerators
12
Multi Tray Aerators
13
Multi Tray Aerators
14
Spray Aerators: Nozzles
15
Spray Aerators: Nozzles
16
Spray Aerators: Nozzles
17
Diffused Air System
18
Diffused Air System
19
Mechanical Aerators
20
Cascade Aerator
Cw,e = gas concentration in the effluent water, mg/L
Cw,0
, = gas concentration in the influent water, mg/L
Cs = gas saturation concentration in water at specific temperature , mg/L
h = height of one step, m
21
Cascade Aerator
22
Cascade Aerator
Ca 
( MW ) gas * C g
Vair * 10 3
Ca
= gas concentration in air, mg/L
(MW)gas = molecular weight of gas, gram
Cg
= gas concentration in air, ppm
i i i
= volume of 1 mole of air at a given pressure and temperature, L
Vair Vair
nRT

P
n = number of air moles, ( in this case n =1)
R = universal gas constant, 0.08285 (atm.L)/(mole.K)
T = temperature in degrees Kelven.
P = air pressure , atm
Typical values of Cg :
CH4 = 1.5‐2.00 ppp
H2S = 20‐30 ppm
CO2 = 350‐450 ppm
23
Cascade Aerator
Example:
Calculate the saturation concentration of H2S in water at air temperature of 20 0C and a pressure of 1 atm if its concentration in air is 30 ppm. Solution:
(MW)gas = 2+32.1 = 34.1 g
2+32 1 34 1 g
1 * 0.08285 * (20  273.15)
 24.29 L
1
34.1* 30
Ca 
 0.042mg / L
3
24.29 *10
From the Table , kD for H
From the Table , k
for H2S at 20 S at 20 oC C = 2.87
2.87
Cs = KD* Ca = 2.87 * 0.042 = 0.1205 mg/L
Vair 
p
p
You can calculate CS for CO2,, O2,, CH4 in the same procedure of this example. 24
Cascade Aerator
Example:
Cascade is to be used to remove methane from ground water. The cascade has C
d i
b
d
h
f
d
h
d h
5 steps each 0.40 m. The initial concentration of methane in the ground water is 0.80 mg/L and its saturation concentration (CS ) in water under atmospheric pressure i i l
is is close to 0 mg/L. t 0
/L
‐What is the efficiency of methane removal after the 5 steps?
‐What is the effluent methane concentration?
Solution:
l i
From the table in a previous slide: kCH4 = 27 % K = 1‐ (1‐k)n = 1‐ (1‐ 0.27)5 = 0.793
Cw,e  0.80
C w ,e  C w , 0
, 0.793 
K 
0.0  0.80
C s  C w ,0
Cw,e = (CCH4) effluent = 0.166 mg CH4/L.
25
Cascade Aerator
Example:
What
h is
i the
h length
l
h off the
h Cascade
C
d aerator in
i the
h previous
i
example
l if its
i capacity
i
is 35 m3/h per 1 m length of the step. The total water flow to be treated is 625 m3/h.
What is the total surface area of the cascade if the width to height ratio of the steps
i 1:1?
is
1 1?
Solution:
 L cascade = 625/35 = 17.86 m
 Acascade = (number of steps ‐1)*step width*Lcascade
= (5
(5‐1)*0
1) 0.40
40*17
17.86 86 = 28.58 m
28 58 m2
26
Multiple
p trayy aerator
This is an economical solution that occupies little space. It consists of three to eight trays with perforated bottoms. The holes are 5 to 13 mm in diameter and spaced at 75 mm center to center.  The trays are placed one on top of another with a 30‐50cm space e ays a e p aced o e o op o a o e
a 30 50c space
between them.  The water is evenly distributed in the first tray through porous pipes or by stopping the gush with a bar (as shown in the figure) so
pipes, or by stopping the gush with a bar (as shown in the figure), so that the water trickles into the tray at a rate of 40‐200 L/min per square meter of the tray’s surface. Small droplets are then sprayed into the next tray
Small droplets are then sprayed into the next tray. To obtain a finer spray, the aerator trays can be filled with thick gravel approximately 10cm deep (50‐150 mm) in diameter. 27
Multiple
p trayy aerator
The removal of carbon dioxide by multiple tray aerators is represented by the following equation ( AWWA,1990):
h f ll
(
)
C
 e  kn
C0
C = effluent concentration, mg/l
C0 = influent concentration, mg/l
k
k = removal rate constant, from 0.28 to 0.37 for CO
removal rate constant from 0 28 to 0 37 for CO2.
n = number of trays 28
Multiple
p trayy aerator
Example:
A ground water containing 8 mg/L of CO2 is to be degasified using multiple tray A ground water containing 8 mg/L of CO
is to be degasified using multiple tray
aerator with three steps. The groundwater flow rate is 990 L/min. The k value
is 0.33 and the hydraulic loading rate of the trays is 400 L/(min.m2)
1 What is the CO
1.
What is the CO2 concentration in the treated effluent?
concentration in the treated effluent?
2. What is the size of the trays if the length to width ratio is 2:1?
Solution:
1.
2.
C = C0 * e –kn = C0 * e –0.33X3 = 8*0.3716 = 2.98 mg CO2/L
Area of each tray = (990 L/min)/[(400L/(min.m2)] = 2.47 m2
Dimensions of the tray : L = 2 W
L* W = 2.47 m2
2W*W = 2.47 m2
W = 1.57 m
L = 3.14
3 14 m
29
Removal of Iron and Manganese from water by Aeration:
 Iron and manganese in excess of 0.30 mg/L and 0.05 mg/L , respectively ,should be removed according to secondary standards of EPA. They cause staining of laundry and plumbing fixtures, produce unacceptable taste and color.
Aeration provides the dissolved oxygen needed to convert the iron and manganese from their ferrous and manganous forms to their insoluble oxidized ferric and manganic forms. The produced precipitates of ferric hydroxide and manganic oxide are then removed by sedimentation followed by rapid sand filtration.
db
di
t ti
f ll
db
id
d filt ti
 It takes 0.14 mg/L of O2 to oxidize 1 mg/L of iron; and 0.29 mg/L of O2 to oxidize 1 mg/L of manganese
of manganese. Oxidation reaction of iron is:
4 Fe(HCO3)2 + 2H2O + O2 > 4 Fe(OH) 3 ↓ + 8CO2
Oxidation reaction of iron is:
2MnSO4 + 2H2O + O2 > 2MnO2 ↓ + 2H2SO4
2Mn(HCO3)2 + O2
> 2MnO2 ↓ + 2H2O + 4CO2
30